3.12g of oxygen is adsorbed on 1.2 g of platinum metal.

Question: $3.12 \mathrm{~g}$ of oxygen is adsorbed on $1.2 \mathrm{~g}$ of platinum metal. The value of oxygen adsorbed per gram of the adsorbent at 1 atm and $300 \mathrm{~K}$ in $\mathrm{L}$ is ________________ $\left[\mathrm{R}=0.0821 \mathrm{~L}\right.$ atm $\left.\mathrm{K}^{-1} \mathrm{~mol}^{-1}\right\rceil$ Solution: (2) Moles of $\mathrm{O}_{2}=\frac{3.12}{32}=0.0975$ volume of $\mathrm{O}_{2}=\frac{\mathrm{nRT}}{\mathrm{D}}=\frac{0.0975 \times 0.082 \times 300}{1}$ $=2.3985 \mathrm{~L}...

Choose the correct answer in each of the following questions:

Question: Choose the correct answer in each of the following questions:The sum of firstnterms of an AP is (4n2+ 2n). Thenth term of the AP is (a) $(6 n-2)$ (b) $(7 n-3)$ (c) $(8 n-2)$ (d) $(8 n+2)$ Solution: LetSndenotes the sum of firstnterms of the AP. $\therefore S_{n}=4 n^{2}+2 n$ $\Rightarrow S_{n-1}=4(n-1)^{2}+2(n-1)$ $=4\left(n^{2}-2 n+1\right)+2(n-1)$ $=4 n^{2}-6 n+2$ $\therefore n^{\text {th }}$ term of the $\mathrm{AP}, a_{n}=S_{n}-S_{n-1}$ $=\left(4 n^{2}+2 n\right)-\left(4 n^{2}-6 n+...

Find the lengths of the medians of

Question: Find the lengths of the medians of a ΔABChaving vertices atA(5, 1), B(1, 5), andC(3, 1). Solution: We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (5, 1); B (1, 5) and C (3,1). So we should find the mid-points of the sides of the triangle. In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as, $\mathrm{P}(x, y)=\left(\fr...

A car tyre is filled with nitrogen gas at 35 psi at

Question: A car tyre is filled with nitrogen gas at 35 psi at $27^{\circ} \mathrm{C}$. It will burst if pressure exceeds 40 psi. The temperature in ${ }^{\circ} \mathrm{C}$ at which the car tyre will burst is________________ . (Rounded-off to the nearest integer) Solution: (70) $\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}$ $\frac{35}{300}=\frac{40}{T_{2}}$ $T_{2}=\frac{40 \times 300}{35}$ $=342.86 \mathrm{~K}$ $=69.85^{\circ} \mathrm{C}=70^{\circ} \mathrm{C}$...

The volume occupied by

Question: The volume occupied by $4.75 \mathrm{~g}$ of acetylene gas at $50^{\circ} \mathrm{C}$ and $740 \mathrm{mmHg}$ pressure is ____________L. (Rounded off to the nearest integer) (Given $\mathrm{R}=0.0826 \mathrm{~L} \operatorname{atm} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ ) Solution: (5) $T=50^{\circ} \mathrm{C}=323.15 \mathrm{~K}$ $\mathrm{P}=740 \mathrm{~mm}$ of $\mathrm{Hg}=\frac{740}{760} \mathrm{~atm}$ $\mathrm{V}=?$ moles $(\mathrm{n})=\frac{4.75}{26}$ $\mathrm{V}=\frac{4.75}{26} \time...

Gaseous cyclobutene isomerizes to butadiene in a first order process which has a

Question: Gaseous cyclobutene isomerizes to butadiene in a first order process which has a ' $\mathrm{k}$ ' value of $3.3 \times 10^{-4} \mathrm{~s}^{-1}$ at $153^{\circ} \mathrm{C}$. The time in minutes it takes for the isomerization to proceed $40 \%$ to completion at this temperature is___________ . (Rounded off to the nearest integer) Solution: (26) For firdst order $\operatorname{Rxn}: t=\frac{2.303}{k} \log \left[\frac{100}{100-x}\right]$ $\mathrm{X}=40, \mathrm{k}=3.3 \times 10^{-4}$ $\ma...

Choose the correct answer in each of the following questions:

Question: Choose the correct answer in each of the following questions: The sum of firstnterms of an AP is (5nn2). Thenth term of the AP is (a) $(5-2 n)$ (b) $(6-2 n)$ (c) $(2 n-5)$ (d) $(2 n-6)$ Solution: LetSndenotes the sum of firstnterms of the AP. $\therefore S_{n}=5 n-n^{2}$ $\Rightarrow S_{n-1}=5(n-1)-(n-1)^{2}$ $=5 n-5-n^{2}+2 n-1$ $=7 n-n^{2}-6$ nthterm of the AP,an=SnSn 1 $=\left(5 n-n^{2}\right)-\left(7 n-n^{2}-6\right)$ $=6-2 n$ Thus, thenthterm of the AP is (6 2n).Hence, the correct...

The number of chlorine atoms in

Question: The number of chlorine atoms in $20 \mathrm{~mL}$ of chlorine gas at $S T P$ is____________ $10^{21}$. (Round off to the Nearest Integer). [Assume chlorine is an ideal gas at STP $\mathrm{R}=0.083 \mathrm{~L}$ bar $\mathrm{mol}^{-1} \mathrm{~K}^{-1}, \mathrm{~N}_{\mathrm{A}}=6.023 \times 10^{23}$ ] Solution: (1) $\mathrm{PV}=\mathrm{nRT}$ $1.0 \times \frac{20}{1000}=\frac{\mathrm{N}}{6.023 \times 10^{23}} \times 0.083 \times 273$ $\therefore$ Number of $\mathrm{Cl}_{2}$ molecules, $\ma...

Choose the correct answer in each of the following questions:

Question: Choose the correct answer in each of the following questions: The sum of firstnterms of an AP is (3n2+ 6n). The common difference of the AP is (a) 6 (b) 9 (c) 15 (d) 3 Solution: LetSndenotes the sum of firstnterms of the AP. $\therefore S_{n}=3 n^{2}+6 n$ $\Rightarrow S_{n-1}=3(n-1)^{2}+6(n-1)$ $=3\left(n^{2}-2 n+1\right)+6(n-1)$ $=3 n^{2}-3$ So, $n^{\text {th }}$ term of the $\mathrm{AP}, a_{n}=S_{n}-S_{n-1}$ $=\left(3 n^{2}+6 n\right)-\left(3 n^{2}-3\right)$ $=6 n+3$ Letdbe the commo...

The pressure exerted by a non-reactive gaseous mixture of

Question: The pressure exerted by a non-reactive gaseous mixture of $6.4 \mathrm{~g}$ of methane and $8.8 \mathrm{~g}$ of carbon dioxide in a $10 \mathrm{~L}$ vessel at $27^{\circ} \mathrm{C}$ is ___________kPa .(Round off to the Nearest Integer). [Assume gases are ideal, $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ Atomic masses: $\mathrm{C}: 12.0 \mathrm{u}, \mathrm{H}: 1.0 \mathrm{u}, \mathrm{O}: 16.0 \mathrm{u}]$ Solution: (150) Total moles of gases, $\mathrm{n}=\mathrm...

Question: The pressure exerted by a non-reactive gaseous mixture of $6.4 \mathrm{~g}$ of methane and $8.8 \mathrm{~g}$ of carbon dioxide in a $10 \mathrm{~L}$ vessel at $27^{\circ} \mathrm{C}$ is ___________kPa .(Round off to the Nearest Integer). [Assume gases are ideal, $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ Atomic masses: $\mathrm{C}: 12.0 \mathrm{u}, \mathrm{H}: 1.0 \mathrm{u}, \mathrm{O}: 16.0 \mathrm{u}]$ Solution: (150) Total moles of gases, $\mathrm{n}=\mathrm...

Solve the following

Question: At $25^{\circ} \mathrm{C}, 50 \mathrm{~g}$ of iron reacts with $\mathrm{HCl}$ to form $\mathrm{FeCl}_{2}$. The evolved hydrogen gas expands against a constant pressure of 1 bar. The work done by the gas during this expansion is______________ $\mathrm{J}$. (Round off to the Nearest Integer) [Given : $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$. Assume, hydrogen is an ideal gas] [Atomic mass off Fe is 55.85u ] Solution: (2218) $\mathrm{T}=298 \mathrm{~K}, \mathrm{R}...

Choose the correct answer in each of the following questions:

Question: Choose the correct answer in each of the following questions: If thenth term of the AP is (2n+ 1) then the sum of its first three terms is (a) 6n+ 3 (b) 15 (c) 12 (d) 21 Solution: nthterm of the AP,an= 2n+ 1 (Given) First term,a1= 2 1 + 1 = 2 + 1 = 3Second term,a2= 2 2 + 1 = 4 + 1 = 5Third term,a3= 2 3 + 1 = 6 + 1 = 7 Sum of the first three terms =a1+a2+a3= 3 + 5 + 7 = 15Hence, the correct answer is option B....

Solve this

Question: If $4, x_{1}, x_{2}, x_{3}, 28$ are in $\mathrm{AP}$, then $x_{3}=$ ? (a) 19(b) 23(c) 22(d) cannot be determined Solution: (c) 22Here,a= 4,l= 28 andn= 5Then, ​T5= 28⇒​a+(n - 1)d= 28⇒​ 4 + (5 - 1)d= 28⇒ 4d=24⇒d= 6Hence,x3= 28 - 6 = 22...

Choose the correct answer in each of the following questions:

Question: Choose the correct answer in each of the following questions: The next term of the AP $\sqrt{7}, \sqrt{28}, \sqrt{63}, \ldots$ is (a) $\sqrt{70}$ (a) $\sqrt{84}$ (a) $\sqrt{98}$ (d) $\sqrt{112}$ Solution: The given terms of the AP can be written as $\sqrt{7}, \sqrt{4 \times 7}, \sqrt{9 \times 7}, \ldots$ i.e. $\sqrt{7}, 2 \sqrt{7}, 3 \sqrt{7}, \ldots$ $\therefore$ Next term $=4 \sqrt{7}=\sqrt{16 \times 7}=\sqrt{112}$ Hence, the correct answer is option $D$....

Prove the following

Question: Let $\vec{b}, \vec{b}$ and $\vec{c}$ be three vectors such that $|\vec{a}|=\sqrt{3}$, $|\vec{b}|=5, \vec{b} \cdot \vec{c}=10$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$. If $\vec{b}$ is perpendicular to the vector $\vec{b} \times \vec{c}$, then $|\vec{a} \times(\vec{b} \times \vec{c})|$ is equal to_______. Solution: $\vec{b} \cdot \vec{c}=10 \Rightarrow|\vec{b}||\vec{c}| \cos \left(\frac{\pi}{3}\right)=10$ $\Rightarrow 5 \cdot|\vec{c}| \cdot \frac{1}{2}=10 \Righta...

Choose the correct answer in each of the following questions:

Question: Choose the correct answer in each of the following questions: The common difference of the AP $\frac{1}{3}, \frac{1-3 b}{3}, \frac{1-6 b}{3}, \ldots$ is (a) $\frac{1}{3}$ (b) $\frac{-1}{3}$ (c) $b$ (d) $-b$ Solution: The given AP is $\frac{1}{3}, \frac{1-3 b}{3}, \frac{1-6 b}{3}, \ldots$ $\therefore$ Common difference, $d=\frac{1-3 b}{3}-\frac{1}{3}=\frac{1-3 b-1}{3}=\frac{-3 b}{3}=-b$ Hence, the correct answer is option D....

If the vectors,

Question: If the vectors, $\vec{p}=(a+1) \hat{i}+a \hat{j}+a \hat{k}$ $\vec{q}=a \hat{i}+(a+1) \hat{j}+a \hat{k}$ and $\vec{r}=a \hat{i}+a \hat{j}+(a+1) \hat{k} \quad(a \in \mathrm{R})$ are complanar and $3(\vec{p} \cdot \vec{q})^{2}-\lambda|\vec{r} \times \vec{q}|^{2}=0$, then the value of $\lambda$ is_________. Solution: $\left|\begin{array}{ccc}a+1 a a \\ a a+1 a \\ a a a+1\end{array}\right|=0$ $\Rightarrow 3 a+1=0 \Rightarrow a=-\frac{1}{3}$ The given vectors $\vec{p}=\frac{2}{3} \hat{i}-\fr...

Choose the correct answer in each of the following questions:

Question: Choose the correct answer in each of the following questions: The common difference of the AP $\frac{1}{p}, \frac{1-p}{p}, \frac{1-2 p}{p}, \ldots$ is (a) $p$ (b) $-p$ (c) $-1$ (d) 1 Solution: The given $\mathrm{AP}$ is $\frac{1}{p}, \frac{1-p}{p}, \frac{1-2 p}{p}, \ldots$ $\therefore$ Common difference, $d=\frac{1-p}{p}-\frac{1}{p}=\frac{1-p-1}{p}=\frac{-p}{p}=-1$ Hence, the correct answer is option C....

Solve this

Question: If $1+4+7+10+\ldots+x=287$, find the value of $x$. Solution: $s_{n}=\frac{n}{2}[2 a+(n-1) d]$ $a=1, d=4-1=3$ and $s_{n}=287$ $287=\frac{n}{2}[2+(n-1) 3]$ $\Rightarrow 574=n(2+3 n-3)$ $\Rightarrow 3 n^{2}-n-574=0$ Using $n=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ $n=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(3)(-574)}}{2(3)}$ $\Rightarrow n=\frac{1 \pm \sqrt{1+6888}}{6}$ $\Rightarrow n=14, \frac{-41}{3}$ Number of terms can not be negative So, $n=14$ $n=14$ $s_{n}=\frac{n}{2}[a+l]$ $287=\frac{14}{2...

Solve the following

Question: $10.30 \mathrm{mg}$ of $\mathrm{O}_{2}$ is dissolved into a liter of sea water of density $1.03 \mathrm{~g} / \mathrm{mL}$. The concentration of $\mathrm{O}_{2}$ in ppm is ______________ . Solution: (10.00) $\mathrm{ppm}=\frac{10.3 \times 10^{-3}}{1.03 \times 1000} \times 10^{6}=10$...

Prove the following

Question: Let $a=\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ be two vectors. If $\vec{c}$ is a vector such that $\vec{b} \times \vec{c}=\vec{b} \times \vec{a}$ and $\vec{c} \cdot \vec{a}=0$, then $\vec{c} \cdot \vec{b}$ is equal to:$-\frac{3}{2}$$\frac{1}{2}$$-\frac{1}{2}$$-1$Correct Option: , 3 Solution: $\vec{a} \times(\vec{b} \times \vec{c})=\vec{a} \times(\vec{b} \times \vec{a})$ $\Rightarrow-(\vec{a} \cdot \vec{b}) \vec{c}=(\vec{a} \cdot \vec{a}) \vec{b}-(\vec{a} \cdot \v...

5 g of zinc is treated separately with an excess of

Question: $5 \mathrm{~g}$ of zinc is treated separately with an excess of (A) dilute hydrochloric acid and (B) aqueous sodium hydroxide. The ratio of the volumes of $\mathrm{H}_{2}$ evolved in these two reactions is:$1: 2$$1: 1$$1: 4$$2: 1$Correct Option: , 2 Solution: $\mathrm{Zn}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2}+\mathrm{H}_{2}$ $\mathrm{Zn}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}$ $\mathrm{NaOH}$ and $\mathrm{HCl}$ reacts with a certain amount of...

Let the volume of a parallelopiped

Question: Let the volume of a parallelopiped whose coterminous edges are given by $\vec{u}=\hat{i}+\hat{j}+\lambda \hat{k}, \vec{v}=\hat{i}+\hat{j}+3 \hat{k}$ and $\vec{w}=2 \hat{i}+\hat{j}+\hat{k}$ be 1 cu. unit. If $\theta$ be the angle between the edges $\vec{u}$ and $\vec{w}$, then $\operatorname{cost} \theta$ can be:$\frac{7}{6 \sqrt{6}}$$\frac{7}{6 \sqrt{3}}$$\frac{5}{7}$$\frac{5}{3 \sqrt{3}}$Correct Option: , 2 Solution: It is given that $\vec{u}=\hat{i}+\hat{j}+\lambda \hat{k}, \quad \ve...

The first and second ionisation enthalpies of a metal are 496 and 4560 kJ

Question: The first and second ionisation enthalpies of a metal are 496 and $4560 \mathrm{~kJ} \mathrm{~mol}^{-1}$, respectively. How many moles of $\mathrm{HCl}$ and $\mathrm{H}_{2} \mathrm{SO}_{4}$, respectively, will be needed to react completely with 1 mole of the metal hydroxide?1 and 12 and $0.5$1 and 21 and $0.5$Correct Option: , 4 Solution: A large difference between first and second ionisation enthalpies $\left(4560-496=4064 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ confirms the metal to ...